Force may be defined as the cause of motion and deformation.
When a force is applied to an object, the object either moves or changes shape
or both. In most cases, it is not possible to detect the deformation by
naked eyes at the molecular or atomic level. Deformation occurs no matter
how small.

In Chapter 1, force was defined as the product of mass and acceleration.
Simply F = Ma. A more useful form of
this formula isΣF
= Ma. ΣF means the sum of
forces acting on mass M. Since forces acting on an object may act in
opposite directions, ΣF is also
called the net force.

The formula ΣF
= Ma is called the "Newton's 2nd Law of Motion."

For example, a car moving along a straight and horizontal highway,
experiences an engine force Fe, while
being opposed by an overall frictional force, Ff (
road friction as well as air resistance). If the car is moving to the
right and to the right is taken to be the positive direction, then
Fe acts to the right and Ff
acts to the left. The net force is
ΣF =
Fe- Ff
.

Example 1: An 850-kg car is
accelerating at a rate of 2.4m/s2 to the right along a straight and horizontal
road where it experiences an overall frictional force of 1500N. Determine
the force that its engine exerts.

Solution: If "to the right" is taken to be
positive as usual, Fe
is positive, and Ff
negative.Note that friction always opposes
the direction of pending motion. Applying Newton's 2nd law:

Note that the 2040N must be rounded to 2 significant figures and then
added to the 1500N.

Example 2: A 2400-kg truck is moving at
a constant speed of 15m/s on a horizontal and straight road that offers an
overall frictional force of 1800N. Calculate (a) its acceleration, (b) the
engine force, (c) the distance it travels in 35s, (d) its acceleration if it
changes its speed to 25m/s in 8.0 seconds, and (e) the engine force in this
case.

Example 3: A car
that weighs 14700N is traveling along a straight road at a speed of 108 km/h.
The driver sees a deer on the road and has to bring the car to stop in a
distance of 90.m. Determine (a) the necessary deceleration, (b)
the stopping force, (c) the brakes force, if the road friction is 2100N, and (d)
the stopping time. Try to solve the problem
yourself before looking at the solution.

Solution: The mass of the car and its velocity in
(m/s) must be determined first. Recall that w = Mg.
The mass of the car is therefore, M = w/g,
or M = (14700N) / (9.8 m/s2) , or
M = 1500kg.

1) If an object is under a zero net
force, it is either stationary or if moving, it moves at constant
velocity. Note that constant velocity means constant speed plus
constant direction that means along a straight line.

2)A nonzero net forceΣF acting on
mass M causes an acceleration a in it such that
ΣF= Ma.
The acceleration has the same direction as the applied net force.

3) There is a reaction for every action,
equal in magnitude, but opposite in direction.

An example covering the 1st and 2nd laws has already been
made (Example 2). For the 3rd law, look at the following
example:

Example 4: A 20.0-kg crate
is on a horizontal and frictionless surface as shown. (a) Calculate and
show the vertical forces acting on this crate. (b) Knowing that the crate
is being pushed to the left by a 53-N force, what magnitude force (F) to the
right must be applied onto the block to give it an acceleration of 2.5m/s2
to the right?

Note that w = 196 N must be rounded to 2 sig. fig. and written
as w = 2.0x102 N. The same is true for F = 103N that
must be rounded to 100 and written as F = 1.0x102 N.

Example 5: An 80.0-kg
man is standing in an elevator. Determine the force of the elevator onto
the person if the elevator is (a) accelerating upward at 2.5m/s2,
(b) going upward at constant speed, (c) coming to stop going upward at a
deceleration of 2.5m/s2, and (d) going downward at an acceleration of
2.5m/s2.

Solution: The force of
the elevator onto the person is nothing but the normal reaction,
N, of the floor onto his
feet. For each case, a force diagram must be drawn. Let's
take the +y-axis to be upward. Make sure that
you carefully draw each force diagram with minimal looking at the following
force diagrams.

(a) w = Mg = (80kg)(-9.8
m/s2) = -780N

This is the case that the elevator has just started going upward.
Since its speed has to change from zero to some value, it has to accelerate
upward and the person feels heavier because the floor of the elevator exerts
a normal reaction, N , onto the
man that is greater than his weight. This creates a nonzero net force
and therefore accelerates the person.

ΣF = Ma
;N- Mg = Ma ;N =M (g + a)

N =M (g + a) =
(80.0kg)(9.8+2.5)m/s2 = 984N

(b) w = Mg = (80kg)(-9.8
m/s2) = -780N

In this case, since the elevator goes up at constant speed, its acceleration
is zero and so is the acceleration of the man. Zero acceleration means zero net force acting on
the man. This requires ( N )
to be equal to (
w ) in magnitude.

ΣF = Ma
;N- Mg = Ma ;N =M (g + a)

N =M (g + a) =
(80kg)(9.8+0)m/s2 = 780N

(c) w = Mg =
(80kg)(-9.8 m/s2) = -780N

In this case, the elevator is coming to stop in
its
going upward. In other words, it decelerates as it goes upward.
We all have this experience that during such slowing down, we feel lighter.
We will notice that the magnitude of the normal reaction,
N, becomes less than that of w.

ΣF = Ma
;N- Mg = Ma ;N =M (g + a)

N =M (g + a) = (80kg)(9.8-2.5)m/s2 =
580N

(d) w = Mg = (80kg)(-9.8
m/s2) = -780N

When the elevator starts going downward, its speed changes from zero to some
value, and therefore it accelerates. This time we use -2.5
m/s2 because the acceleration vector is downward.

ΣF = Ma
;N- Mg = Ma ;N =M (g + a)

N =M (g + a) = (80kg)(9.8-2.5)m/s2
= 580N

Example 6: In the figure shown,
determine the acceleration of the system of blocks:

Solution: Does free fall occur? If the cord is broken,
does free fall occur? The answer to the 2nd question is definitely "Yes."
The answer to the first question is now more clear, "No." Block B is not
free to fall and must pull block A in addition to moving itself. The
vertical forces on block A cancel each other, according to the Newton's 3rd law.
The force of gravity on block B is the cause of motion while block A is on the
horizontal surface. In fact, we are only interested to find the
acceleration of the system of blocks while A slides horizontally. Let's
summarize: (1) the force that causes motion is WB and (2) this force
has to move both masses (MA + MB). Since the system
is connected, both blocks move at the same acceleration (same magnitude). We
may write:

ΣF = Ma
;wB =
(MA + MB)
a; 29N = ( 5.0 kg
+ 3.0 kg) a;a = 3.6 m/s2

Example 7:

Vehicles A and B are shown in nine different
cases. In each case, a statement is made on the left. Refer
to the figure in the middle and determine if the statement is true (T) or false (F).

5) ΣF
= Ma implies (a) the proportionality of the net force and the
acceleration it generates in mass M (b) the proportionality of the net force and
mass (c) both a and b.click here

6) If the same force is applied to masses M1 and M2
separately, knowing that M1>M2, then (a) M1
accelerates greater than M2 does (b) M1 accelerates
less than M2 does (c) both M1 and M2
gain the same acceleration.click here

7) The acceleration of gravity on an object in the vicinity of
the Earth (a) is the same whether the object is stationary or falling freely
(b) acts on the object only if it is falling (c) acts on the object when
it is placed on the horizontal surface only.

8) The acceleration of gravity (a) is exactly zero where the
Space Station is (b) is not exactly zero where the Space Station is, but
its direction is different there (c) does not completely diminish
with distance.click here

9) The direction of the acceleration (and hence, force of
gravity) on an object around the Earth (a) always passes through the center of the
Earth (b) is perpendicular to the local free water surface (c)
both a and b.

10) The direction of g , the gravity
acceleration, at any location is (a) the same as the plumb-line at that
location (b) is a few degrees different from the plumb-line at that
location (c) is normal to the plum-line at that location.click here

11) According to Newton's 1st law (a) net force and mass
are proportional (b) a moving object under a zero net force has a
constant speed only (c) a moving object under a zero net force has a
constant velocity.

12) When you are pushing a shopping cart horizontally and at a
constant velocity, your force on the cart is (a) greater than the force you feel
from the cart on your hand (a) is less than the force you fell from the cart on
your hand (c) equal to the force you feel from the cart on your hand.click here

13) When you are accelerating a shopping cart horizontally, your
force on the cart is (a) greater than the force you feel from the cart on you
(a) is less than the force you fell from the cart on you (c) equal to the
force you feel from the cart on you.

14) When you are slowing down a shopping cart horizontally, your
force on the cart is (a) greater than the force you feel from the cart on you
(a) is less than the force you fell from the cart on you (c) equal to the
force you feel from the cart on you.

16) Is the direction of ΣF in ΣF
= Ma,
the same as the direction of acceleration (a)? Ans:
...... click here

17) If the net force acting on a mass (M) is not zero, can that
mass move at constant velocity? Ans.: .......

18) If the net force on an object is zero, is the object
definitely moving at constant velocity? Ans.: .......

19) A box weighs 35N and is placed on a horizontal surface.
Is the normal force 35N? Ans.: ...... click here

20) A 75-N box is placed on a horiz. surf. with a boy
sitting on it. Is the normal force on the box 75N? Ans.: ......

21) When an object is accelerating horizontally on a horizontal surface, is the
horizontal external force on the object equal to the horizontal friction force of surface on the object?
Ans.: ...... click here

22) Does a lady standing in an elevator accelerating
upward feel heavier because the normal force from the floor on her is greater than her
weight? Ans.: ......

23) A spaceship in outer space is moving at constant velocity of
55 miles/sec. An astronaut looking through a window notices a bolt
becoming detached from the ship. How many miles behind the ship will the
bolt be in an hour? Ans.: ...... click here

24) A spaceship moving along a straight path wants to change its
speed from 55 mi/s to 75mi/s at a relatively low acceleration of (1/3)g to keep the occupants
in some relative comfort. How long will it take? How many miles does
it travel for the change to happen. Ans.: ......

25) If the spaceship in (Question 24) weighs 490,000N on Earth,
what average force should its jets exert on it during the acceleration period?
Ans.: ...... click here

Friction:

Friction is the result of
engagement of surface irregularities between two surfaces in contact.

Coefficient of Kinetic Friction ( μk
) :On a horizontal surface, the
ratio of the horizontally applied force (Fappl.)
to an object to the weight of the object (w), to slide the object at
a constant velocity, is called the coefficient of kinetic friction. This is
mathematically written as:

Note that Fappl.
= Fk and w = N .

It is easy to understand that Fk
( force of kinetic friction) is the force that the horizontal surface exerts on
the object and Fappl.
is the force that equals Fk. If
Fappl. is exactly equal to
Fk, the object slides at constant velocity.
We also know that if w and
N are the only forces in
the vertical direction, then
N = w ; therefore, the above formula may
be written as

This formula is often written in its cross-multiplied form:Fk = μkN. This means that, if the coefficient of friction between two
surfaces is known, and we can also determine the normal or compressive force
between those two surfaces, we are then able to determine the necessary force
that can slide one surface against the other at a constant velocity.

Example 8: The coefficient
of kinetic friction between a cement block and a plank of wood is 0.38.
The block has a mass of 15kg and is placed horizontally on the plank. Find
the magnitude of the horizontal force that can push the block to the right at
a constant velocity.

Example 9:

In the figure shown, determine the magnitude of the horizontal
force to the right that can move the block at (a) constant velocity, and (b) at an
acceleration of 3.0 m/s2.

Solution: (a)

w = Mg = (25kg)(9.8m/s2) = 245N

N =
245N ;Fk = μkN
= (0. 26)(245N) = 64N

ΣF = Ma ; F -
64 = (25)( 0 ) ;
F = 64N

Solution: (b)

ΣF = Ma ;
F - 64N = (25)(3. 0)N ; F = 139N

Coefficient of Static Friction ( μs
) : On a horizontal surface, the
ratio of the horizontally applied force (Fappl.)
to an object to the weight of the object (w),
to bring the object onto the verge of slipping, is called the coefficient of
static friction. This is
mathematically written as:

Again,
N = w ,
and Fappl.
= Fs ,on the verge of
slipping, and the above equation may be written as:

This formula is often written in its cross-multiplied form:Fs = μsN. This means that, if the coefficient of static friction between two
surfaces is known, and we can also determine the normal or compressive force
between those two surfaces, we are then able to determine the necessary force
that can bring the object onto the verge of slipping.

Example 10: The coefficient
of static friction between a cement block and a plank of wood is 0.46. The
block has a mass of 15kg and is placed horizontally on the plank. Find the
magnitude of the horizontal force that can bring the block onto the verge of
slipping.

Friction Laws:

There
are 5 laws for friction. The first 3 apply to the force of friction, and
the last 2 to the coefficient of friction.

4) Coefficient of friction (μ
) depends on the materials of the contacting surfaces.

5) Coefficient of friction (μ
) depends on the smoothness of the contacting surfaces.

Example 11: Two kids are
sitting on the opposite sides of a 3.0-m long table and sliding a 150-gram
empty cup toward each other, back and forth. The game is to give the cup the right initial
velocity at one edge such that it comes to stop exactly at the opposite edge as
shown.
The diameter of the cup is 10.0cm. The coefficient of kinetic friction
between the cup and the horizontal tabletop is 0.12. Determine the
necessary initial speed.

Solution: Time is not given. At a
first glance, you may think that ( Vf2- Vi2
= 2ax ) is a good idea. In this equation, the
acceleration (a) of the cup is not known. This
means that we need to use the kinetic
equation,
ΣF= Ma in order to solve for acceleration, first.
What are the forces acting on the cup after it is given an initial instant push?
The only acting force is the force of kinetic friction, Fk.
To find Fk, we
need to knowN, and
consequently (w). Therefore, we must start from
w
= Mg.

w = Mg = (0.15kg)(9.8 m/s2)
= 1.47N ; therefore,
N =
1.47N.

Fk = μkN
; Fk =
(0.12)( 1.47 N ) = 0.176 N.

ΣF
= Ma ; - 0.176N = (0.150kg)(a)
; a = -1.173 m/s2.

Vf2- Vi2
= 2 a x ; 02- Vi2
= 2 (-1.173 m/s2)(2.9m)
;Vi = 2.6 m/s.

Example 12: A truck that
weighs 29,400N traveling at 72.0 km/h on a horizontal and straight road skids to
stop in 6.00s. Determine (a) its deceleration, (b) the
stopping force, (c) the kinetic coefficient of friction between its tires
and the horizontally straight road, and (d) the stopping distance.Important: First draw a diagram for the problem and show
all forces acting on the truck. g = 9.8m/s2.

Example 13: A 12-kg
box is placed on a horizontal floor for which
μs =
0.43 and μk =
0.33. Does a 57-N force, applied horizontally to this box, put it into
motion? If yes, will the motion be accelerated or at constant speed?
If accelerated, how far will it travel in 3.0s?

Solution: We know that
Fs > Fk.
If Fs(the force of static friction) is less
than 57-N, the horizontally applied force, motion will occur. Let's
calculate Fs.

1) Friction is the result of the (a) engagement of surface
irregularities between two contacting objects (b) the molecular
attraction between two contacting objects (c) both a and b.click here

2) The importance of normal force, N, between two contacting surfaces is that (a) it helps us calculate the
force of friction (b) it is a measure of the extent two objects (in
contact) push against each other (c) both a and b.click here

3) It takes 12N to horizontally push a 45-N block, placed on a
horizontal surface, at constant velocity. Another 45-N block is placed on
the top of the 1st one. To push both blocks at constant
velocity, it takes a force of (a) 12N (b) 24N (c) 36N.

4) Coefficient of friction,
μ , is (a) the ratio of force
of friction, Ff, to
the normal force, N (b) the ratio of the horizontally applied force, Fappl.,
to the weight force, w, when the object is on a horizontal
surface moving at a constant velocity (c) both a and b.

6) On a horizontal surface, for a single block of weight
w, and a horizontally applied external force,
Fappl., on the block, if the block is sliding
at constant velocity, we may write: (a) N
= w (b)Fappl.= Ff
(c) both a and b. First draw a force diagram, then solve and
answer.click here

7) On a horizontal surface, it takes a horizontal force of (a)14N (b)22N (c)11N to push a 55-N block at
a constant velocity,
knowing that the coefficient of kinetic friction between the block and the surface is 0.40.
First draw a force diagram, then solve and answer.

8) On a horizontal surface, it takes a horizontal force of (a)98N (b)22N (c)11N to push a 50-kg block at a constant velocity,
knowing that the coefficient of kinetic friction between the block and the surface is 0.20.
First draw a force diagram, then solve and answer. click here

9) First draw a force diagram, then solve and answer. To
accelerate a 25-kg block horizontally on a horizontal surface at a rate of 4.0m/s2
, knowing that
μk = 0.30, a horizontally applied force of (a)173.5N (b)73.5N
(c) .5N is needed.

10) First draw a force
diagram, then solve and answer. To accelerate a 50-kg block horizontally
on a horizontal surface at a rate of 4.0m/s2 , knowing that
μk = 0.30, a horizontally applied force of (a)347N (b)473.5N (c)273.5N is needed.

11) In the
figure shown, the applied force, F, is not
horizontal, we may write:

15) Redraw the figure of Question
11 with F = 140N , M = 23kg, and μk = 0.30 and calculate the acceleration of the
block in its sliding to the right. The acceleration is (a) 2.42m/s2
(b)1.42m/s2 (c) 5.42m/s2.click here

16) Redraw the figure of Question
11 with F = 110N , M = 33kg, and μk = 0.150 and calculate the acceleration of the
block in its sliding to the right. The acceleration is (a)1.167m/s2
(b)1.47m/s2 (c)1.38m/s2.click here

17) The coefficients of kinetic and static friction for a block
and surface are 0.26 and 0.38 respectively. If the block
weighs 120N and the surface is horizontal, can a 35N horizontal force put the
block into motion? .......... click here

18) The coefficients of kinetic and static friction
for a block and surface are 0.36 and 0.54, respectively. If
the block weighs 325N and the surface is horizontal, can a 205N horizontal force
put the block into motion? .......... click here

19) If the answer for Question 18 is "Yes", at what acceleration
will the block slide? .......... click here

20) The coefficient of friction,
μ,
between two surfaces with constant characteristics (a) is proportional to the applied force
(b) is a constant (c) depends on the materials of the contacting surfaces
(d) b and c.click here

21) The force of friction, Ff,
(a) is proportional to the applied force (b)
is a constant (c) does not depend on the material of the contacting
surfaces (d) is proportional to the normal force, N.click here

22) The coefficient of friction,
μ, (a) depends on the roughness or smoothness of
the contacting surfaces (b) is a constant (c) depends on the
material of the contacting surfaces (d) all of the above.click here

23) Suppose that you have a bucket of water and a string just
strong enough to hold it hanging. What happens if you try to lift the
bucket by a few inches? Does it (a) pull? or (b) break?

24) Refer to Example 6 and replace the 3.0-kg hanging
block by a 5.0-kg one. In the absence of friction, the
system of blocks move at an acceleration of (a) 9.8m/s2
(b) 4.9m/s2 (c) 2.5m/s2.click here

25) Draw a force diagram for the hanging block of Question 24
and show the 49-N force, the block's weight, as well as the unknown tension,
T, of the cord.T should
be shown as an upward force vector on the top edge of the square you draw as the
block. Since you know the acceleration of the block from Question 24, you
can apply
ΣF = Ma
just to this block and solve for T, the tension in the
cord. We know that the block accelerates downward at 4.9 m/s2
as you have already solved. If you assume upward to be positive, then
T is positive, the 49N is negative and so is the
acceleration. Apply
ΣFy = May
and solve for T. The answer is (a)17N
(b) 24.5N (c)73.5N.

Problems:

1) A car traveling at a constant velocity on a horizontal and
straight road is facing air resistance as well as a road frictional force of
2400N. The engine is exerting a force of 4200N. Calculate the
force due to air resistance.

2) In the figure shown, the 3.0-kg
hanging block causes motion of the two connected blocks at constant
velocities. Find the coefficient of friction between the top block
and the horizontal table.

3) In the figure
shown, find the coefficient of friction if the acceleration of the motion
of the system of blocks has a magnitude of 0.75m/s2.

4) A 2.0-kg rock is hanging from a rope that can
withstand a maximum tension of 33N. (a) If the rope is pulled upward at an
acceleration of 5.0m/s2, will the rope break? (b) Will
it break if the upward acceleration given to the rope is 7.0m/s2?
Support your answers with complete calculations.

5) A 55-kg lady is standing in an elevator.
Assuming g = 9.80 m/s2, calculate the
force of elevator on her feet for the following cases: the
elevator is (a) at rest, (b) moving downward at an acceleration of 3.0m/s2,
(c) moving down at a constant velocity, and (d) is coming to stop in its
downward motion at a deceleration of 3.0m/s2.

6) An 800.-kg car traveling at a velocity of 25 m/s
eastward, comes to stop within a distance of 50.0m. If the brakes
apply a force of 3800N, find the road's overall frictional force.

7) A train moving at 30.0 m/s resumes this speed for 25.0
minutes. Find (a) its acceleration and (b) the distance it travels during
this period. It then slows down to 20.0m/s within a distance
of 125m. Find (c) its deceleration, and (d) the elapsed time during the
slowing down phase.